METHODS FOR FINDING FACTORS OF LARGE INTEGERS*

Size: px

Start display at page:

Download "METHODS FOR FINDING FACTORS OF LARGE INTEGERS*"

Elizabeth Tate
5 years ago
Views:

1 542 H. s. VANDiVER [Nov.-Dec, ] <$ 4^Jsin 2 -^ = 2 < 2'( cosc^) = 2 (m mn), and r x has the value But this reckoning does not seem to lead to any particularly simple geometric interpretation for r. It may turn out, in some circumstances at least, that r t is the more significant measure after all. THE UNIVERSITY OF MINNESOTA METHODS FOR FINDING FACTORS OF LARGE INTEGERS* BY H. S. VANDIVER. Introduction. We shall examine, in this paper, the problem of finding factors of integers beyond the range of Lehmer's factor tables, by methods shorter than that of dividing the integer by all the primes less than its square root. Three methods will be proposed here. The first two depend on the representation of the integer as a definite quadratic form, and the third on the representation as an indefinite quadratic form. As I hope to devote another paper to the development of the last two methods, only outlines and a few examples will be given in connection with them. The theory of quadratic forms has been applied in several different ways to the problem, t In particular, Seelhoff * gave an expeditious method with the use of tables, which, however, is limited in application, * Presented to the Society, September 7, 923, under the title A method of finding factors of integers of the form 8 n +. The author was enabled to carry out this investigation through a grant from the Heckscher Foundation for the Advancement of Eesearch. f Dickson, History of the Theory of Numbers, vol., pp % AMERICAN JOURNAL, vol. 7, p. 264; vol. 8, p, 26.

2 924.] FACTORS OF LARGE INTEGERS 543 since if the integer is composite, it does not always enable us to determine the factors. The first method explained in the present paper applies only to integers of the form 8w+ and 2^+ which are the product of two distinct primes. Although the process will always yield the factors in a finite number of steps, it is impracticable on account of its length if the integer is as large as 0 7. Unlike other methods, however, the particular forms used are always the same. Hence they lend themselves conveniently to the construction of tables with a view of lessening the amount of computation involved. The theory of indefinite forms has been applied by Tchebycheff * to our problem, but the method is quite different from the one given here which involves indefinite forms. 2. First Method of Factorization. Suppose that we have an integer m of the form 8 n + which is the product of two distinct prime factors. It is immediately seen that each factor has the same residue modulo 8, otherwise the product would not be of the form 8n + l. If both are of the form 8n + l or both of the form 8w + 5 then m maybe presented in more than one way in the form x 2 + y 2, x and y prime to each other. Similarly if both are of the form 8w + 3 then m may be represented in more than one way in the form x 2J r 2y 2, x and y prime to each other. In the case where both are of the form 8n + 7 it will now be shown that m may be represented in at least two different ways in the form 2x 2 y 2, where x and y are each <Vm and prime to each other. In a paper which appeared in this BULLETIN (vol. 22 (95), pp. 6-66) I proved that if p is a prime and a is a positive integer prime to p, then there is at least one and not more than two sets (x, y) such that ay = ± # (modjp) when x and y _are positive integers prime to each other and Q<x<cVp, 0<y<]fp. * LIOUVILLE, (), vol. 6, p. 257.

3 544 H. s. VANDIVER [Nov.-Dec, On page 64 of the same article, I remarked that the existence of at least one set (x, y) satisfying the conditions () ay = Azx (modm) where m is composite followed also from the reasoning used. If two sets (x x y t ) and (x 2 y 2 ) are regarded as the same if and only if x x y 2 = x 2 y x then for m any integer there are not more than two sets, a fact which was pointed out to me by Dr. C. F. Gummer. If J kxi = 2/ (modm), kx 2 = y% (modm), where #i, \yx\, \x 2 \ and \y 2 \ are each < Vp, then from the first congruence kx x x 2 = x 2 y x (modm); and from the second y 2 x± x 2 y ± = 0 (modm). This gives, on account of the range of values for the x's and?/'s, y 2 Xi x 2 yi = ± m or y 2 x x x 2 y t = 0, and the last relation can exist only if we regard the sets (x ±, y x ) and (a*, y%) as the same. If the first relation holds and therefore the two sets exist, then a third set different from them cannot exist. To show this we note in (2) that we may take both x x and x 2 positive, and if this is done then it follows, using the relation y 2 x x x 2 y x = ±m, that y t and y 2 have opposite signs. Hence we may write kx x = y x (modm), kx s = 2/3 (modm), where x x and x s are positive and y x and y B have the same sign. It follows as before that ysxt xsyi^-- ±m, and this is impossible since the expression on the left is, in absolute value, less than m.

4 924.] FACTORS OF LARGE INTEGERS 545 Now consider the solution of (3) u 2 2 = 0 (modm), where m is the product of two distinct prime factors each of the form 8n-{-7. There are four incongruent roots of the congruence which we will designate by a, a, b, b, where a = E ±_ b. According to the theorem first proved we may write av = w (modm), iwi = wi (modm), where v, w, v ±, iv ± are each less than Y m. Squaring and using (3) we have (, J 2v 2 iv 2 = 0 (modm), 2v 2 wl = 0 (modm), or, 2v ivi = 2 iw (modm), and since m is odd, (5) (VW-L viw)(vwi~\-viiv) = 0 (modm). Also from (4) we have (6) \ 2 /r^r m ' [ zvi wi = m. We shall now show that in (5) one of the factors on the left is divisible by a factor of m but not by m. For if w0i + #iw = O (modm), then, since v, w, v i9 w x are all prime to m, we have, from (4), since m is not divisible by a square, = (modm), W Wi or a = b (modm), contrary to hypothesis; and if vwi ViW = 0 (modm), then a = b (modm), which is also contrary to hypothesis.

$If a number m of the form 8^ + is the product of tivo distinct primes then it is expressible in at least two different ways by one of the forms x 2 -\-y 2, x 2 -\-2y 2, and 2y 2 x 2, where x and y$

5 546 H. s. VANDivER [Nov.-Dec, Consequently, by taking the smallest positive factor in (5), and finding the greatest common divisor of it and m, we obtain a factor of m. THEOREM. If a number m of the form 8^ + is the product of tivo distinct primes then it is expressible in at least two different ways by one of the forms x 2 -\-y 2, x 2 -\-2y 2, and 2y 2 x 2, where x and y are each positive and <Km. It is also possible to find tivo such representations which yield, by a direct process, a factor of m. This theorem gives the following scheme for finding a factor of an integer of the form 8n-\-l. (.) Divide the integer by all primes less than its cube root. If it is divisible by a prime within this limit which is not of the form 8w+l, the process cannot be carried further. Otherwise w T e may proceed as follows. (2.) Extract the square root of the possibly new number m. If it is a perfect square, all the factors of m are known, as it cannot have more than two distinct prime factors other than unity. If m is not a perfect square we proceed to the third step. (3.) Find by Gauss' method of exclusion all representations of m in the form x 2 + y 2. If there is but one representation, then m is prime. If there is more than one, then the factors of m may be found from the several representations. If there are no representations then proceed to the next step. (4.) Find by the method of exclusion all the representations of m in the form 2y 2 x 2, where x and y are each < Vm. It cannot have a unique representation, as it follows from this that m is prime, and this would have been detected in step 3. If there is more than one representation then the factors of m may be found from them. If there are no representations then we use the fifth step. (5.) Find by the method of exclusion all the representations of m in the form x 2 -\-2y 2. The factors of m are obtained from the several representations.

6 924.] FACTORS OF LARGE INTEGERS Example. Let us show how to factor = N. Divide by all the primes < 82, since [VN] = 8. It is found that N is not divisible by any of these primes. Also 2Vis not a square. Hence it is prime, or is the product of two distinct primes. We seek first all the values x, y, such that x 2 + y 2 = N. Either x or y is odd* and each is <730. Suppose x is odd. Also N=2 (mod 3). This gives x = + (mod 3), since if x = 0 (mod 3) we find y 2 = 2 (mod 3), which is impossible. Hence since x is odd, we have x = ± (mod 6). Now N= (mod 5); and, in a similar way, we find x = 0, + (mod 5). Therefore x =, 5,,9, 25, 29 (mod 30). We have N= 5 (mod 7), and x = ±, ±2 (mod 7). Modulo 20, x has the possible residues, 5, 9, 29, 4, 55, 6, 65, 7, 79, 89, 2, 25, 3, 39, 45, 49, 55, 69, 8, 9, 205, 209. Similarly x has the possible residueso, ±, ±2, ±5 (modll); and we then set down the least possible residues of x modulo 20, which are less than 730. We find 68 numbers. Using the modulus 6, we may exclude numbers of the form 3, 5, and 3, modulo 6. Similarly we may exclude numbers of the form, 4, 6,, 4, 9, 2, 24, modulo 25, and 3, 4, 5, 0,, 2, modulo 3. This leaves 2 possible values for x which we may test directly, and we find none that will give a representation of N in the required form. We now proceed to step 4 and find all representations of the form _ 9 9, T where x and / are each < VN. In this relation 56 < y < 730. Also y = (mod 2). Possible values of y satisfy y = 0, ± (mod 5), y = 0, + (mod 7). Modulo 70, y has therefore the possible residues, 5, 2, 29, 35, 49, 55, and 69. We find similarly y = 0, ±3, ±4 (mod 9). We then write down the integers between 56 and 730 having the above properties and find 59, 525, 545, 56, 58, 589, 609, 65, 645, 65, 679, 699, 75. * The work could have been made shorter here by taking x even and noting that x = 0 (mod 4). 35*

7 548 H. s. VANDIVER [Nov.-Dec, Now y cannot be congruent to 0, 4, 5, 6 or 7 (mod ), and we may exclude integers of this type from the above set, leaving 59, 525, 58, 65, 65, 679. Actual trial of all these gives only two representations, = =,N, and = The greatest common divisor of N and is 823. Hence N= , each of which is a prime. A similar method may be applied to integers of the form 2n+. If -A^is of this form and the product of two distinct primes, then each has the same residue, modulo 2. If both are of the form 2w + l, or 2^ + 5, then Nis expressible as the sum of two squares. If each is of the form 2w + 7 then N = x 2 + 3?/ 2. If each is of the form 2n + then it may be_shown that 2N = 3y 2 x 2, where x and y are each < VN. For in case the congruence a 2 3 = 0 (mod N) is solvable and if a is one of its roots then rna = k (mod N) where m and k are each < VN, and therefore 3m 2 k 2 = 0 (mod N) and 3m 2 k 2 = N or 2JV". But the first case is impossible, since N= (mod 3). We then have all the material necessary to carry through a method for factoring integers of the form 2n-\-l analogous to that described for the case 8w+l. 4. Second Method of Factorization. The second method of factorization is briefly as follows. Find by known methods a negative integer a which is a quadratic residue of n, the integer to be factored (n not a perfect power). From a known result in the theory of quadratic forms it follows that, corresponding to every root u of the congruence u 2 = a (modn) there is a set of. integers x, y and k such that () x 2J r ay 2 = kn, k< 2 Y a/3, x and y prime to each other.* From () * This theorem was used hy H. J. S. Smith, WOEKS, vol., p. 48 in the solution of the quadratic congruence.

8 924.] FACTORS OF LARGE INTEGERS 549 we have ( a/h) = where Jc t is any divisor of Tc, and (W«i) = where a x is any divisor of a. We select the fc's satisfying these conditions, and also &< 2 Va/3 and we find the possible forms a%+ay* = Jc 8 h, s =, 2,..., i. Find by Gauss' method of exclusion or other schemes all actual representations of this type ; then all factors of n may be found from them, by known methods. It is best to select a, if possible, so that it contains several small primes as factors since this will diminish the number of possible values of Jc. 5. Example. n = In order to find values of a as described above, expand Vn as a continued fraction. We have the corresponding values given in the following table. Denominator of comp. quotient 040 = = 4 2, = = = Terms in cont. fraction Hence 5-3 is a residue of n, 9-2 is an E (residue), 5-9 is an B, 5 is an JB, 0 is an JS, whence 2 and 3, 9 and 5 are residues. Also 79 is an JS. Hence a = is a residue. Hence x 2 -\-ay 2 = Jen, where h < 8. From the relations p* nql=azd r where p r /q r is a convergent and d r a denominator of comp.

9 550 H. s. VANDIVER [Nov.-Dec, quotient in development of Vn we have (n/3) = (w/79) = {nib) =. Hence also (ft/5) = (ft/3) = (ft/79) =. This gives ft = 0,, 4 (mod.5) and ft = 0,, 3, 4, 9,0,2 (mod 3). We set down the integers We note also that fte^ 0, 3, 4, 5 (mod 8); using this with ft ^ 2 or 3 (mod 5) we have remaining We have \ = / \ == _ \ 3 / \ 7 Hence ft cannot be a multiple of 3 or 7. Moreover, it is evident that ft is not divisible by 5 2. This leaves From x 2 + ay* = 0% we have 5 x y 2 2w whence (3-79/5) = which is incorrect; this excludes 0. Similarly 26 is excluded, also 65. We have (74/79) = = (94/79) and these are then excluded leaving, 55, 79 and 95 as possible values of ft. Let us now examine x 2 + ay 2 n. Here y ranges from to 7 inclusive, and it must be even, which gives us = n.

10 924.] FACTORS OF LARGE INTEGERS 55 For k = 55, we have hx y 2 = lln. Gauss' method of exclusion we have y = ± 2 (mod 5), y = l (mod 2), whence y E E 0 (mod 3), y=t=±2(mod7), = lln and the two representations give n = Using 6. Example 2. w = (an 8-flgure number selected at random, except that it is not divisible by small integers). Denominator of complete quotient 2743 = = = = 5709 J 2 93 = 386 Term of cont. fraction Hence we may take a = = 806, () x 2 + ay 2 = nk, k<2 VöM or k< 04. Also (w/3) = {nil) = (n/93) = and (ft/3) = (ft/7) = (fc/93) = l. ^ = 0 or (mod 3), ft ^0,, 2, 4 (mod 7); hence, modulo 2, the residues of ft are, 4 ; 7, 9, 5, 6, 8, 2, so the values of ft may be written ( Also ft ^ 0, 4, 5, 7, (mod 8). Using these conditions together with the methods employed in the last case gives ft=, 25,42,43,46, 67.

11 552 H - s. VANDivER [Nov.-Dec, Hence the factors of n may be found by using these values of ft in (2). For the same value of n, n = , we may take a = 3-2. This gives ft =, 3, 6, 43, 49, 53, 56. Letn = Herew = ; Let us now examine x y 2 = kn. We find ft =, 25, 23 or 38. Let n = , the numerator of the 2th Bernoulli number; then n = , Jen = x y 2, whence, using previous methods, ft=l, 6, 3, 39 or Case of Periodic Continued Fractions. In applying the schemes outlined in the second method we may find at the beginning that, after a few operations, we have developed a period of the continued fraction for Vn. Suppose* that the period has 2r terms, say /*l? ^2 i 7 ftr ii v> f*r > ftr 2?? Ply then d r, the denominator of the complete quotient corresponding to & is a divisor of 2n. Since b is not the last quotient in a period then ( lydr^l, and therefore ( l) r d r =, d=2, or some factor of n not unity. Also, if the period has an odd number of terms then x 2 ny 2 = has solutions. It then follows that either a factor of n has been found, or else n is expressible in one of the forms x2j ry 2? x 2 -\-2y 2 or 2x 2 y 2, x and y each <V r p. In cases where the factor is not derived as a divisor of d r, however, it is more expeditious to expand the square root of some multiple of n as a continued fraction to find suitable values for a. * Mârcker, ORELLE, vol. 20 (840), pp c?

12 924.] FACTORS OF LARGE INTEGERS Example. Let us factor n = A period in the development of Vn is,, 2,,, 2056, and d s = 409, whence n = Third Method of Factorization. The third method depends on representing the number as an indefinite quadratic form. In a paper already cited I proved that there exists at least one set of positive integers (x, y) such that ay^^zx (modm) where a is any integer prime to the integer m, x and y each < Vm. Hence if there exist roots of the quadratic congruence p 2 = a (mod n) then corresponding to each root ^? we have jt set {x, y) such that fay = dt x (mod ri), x and y < Vn, and we have ay 2 x 2 = 0 (mod n) or ay 2 x 2 = kn where 0< k< a. 0. Example. Let us consider n = The development of Vn as a continued fraction (given above) shows that 2 is a quadratic residue of n, and therefore 2rr 2 y 2 = kn where 0<&<2, also x and y<vn. The continued fraction development gives (nil) =, (n/3) =, hence (&/7) =, (fc/3) =. Using (k/s) = we have as possible values of k We have k = =, 2, 4 (mod 7) and k E = 0, 2, 6 (mod 8), which leaves 3, 5, 2 and 7 as possible values. Note that V(k+l)n/2 >x> VknJ2Ï. Since writing what precedes, I have examined Kraïtchek's Théorie des Nombres (Paris, 922). This book contains tables that are admirably adapted for use in connection with any of the three methods described in this paper. Kraïtchek tabulates the incongruent values of x in x 2 -{-Dy 2 = N (mod^), where Q assumes various small values. CORNELL UNIVERSITY

2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer?

2x 1 7. A linear congruence in modular arithmetic is an equation of the form. Why is the solution a set of integers rather than a unique integer? Chapter 3: Theory of Modular Arithmetic 25 SECTION C Solving Linear Congruences By the end of this section you will be able to solve congruence equations determine the number of solutions find the multiplicative